Question

Given `f(x)=x^3+2` and `g(x)=root3(x-2)` how do you show they are inverses?

Answer 1

It's not elegant, and I'm sure someone will come along soon with a more-elegant demonstration, but one way would be to substitute in, say, -2, 0 and 2 for x in f(x), find the solution, then substitute that solution into g(x) and see whether it yields the original number. Also the reverse.

Explanation:

Completing f(x) then g(x):

For -2:

`f(-2)=(-2)^3+2=-6`
`g(-6)= root(3)(-6-2) =-2`

For 0:

`f(0)=0^3+2 = 2`
`g(2) = root(3)(2-2)=0`

For 2:

`f(2)=2^3+2=10`
`g(10)=root(3)(10-2)=2`

Completing g(x) then f(x):

For -2:

`g(-2)=root(3)(-2-2)=-1.5874`
`f(-1.5874)=(-1.5874)^3+2=-2`

For 0:

`g(0)=root(3)(0-2)=-1.2599`
`f(-1.2599)=(-1.2599)^3+2=0`

For 2:

`g(2)=root(3)(2-2)=0`
`f(0)=0^3+2=2`

Answer 2

See explanation. Short answer: switch `x` and `y` and solve for `y`.

Explanation:

To find the inverse of a function, simply switch the `x` and `y` variables and solve for `y`.

Let's apply this concept to the first function given and see if we end up with the second function. Also, note `f(x)` is basically `y`.

`f(x) = x^3 + 2`

`y = x^3 +2`

First switch the variables.

`x = y^3 + 2`

Now we need to solve for `y`, so isolate `y`.

`x-2 = y^3 cancel(+2-2)`

`x-2 = y^3`

`root(3)(x-2) = root(3)(y^3)`

`root3(x-2) = y`

We have solved for `y`. Now let's set it equal to the second function and determine if they match.

`root3(x-2) = root(3)(x-2)`

The two functions are identical. Therefore, they are inverses.

Answer 3

See explanation...

Explanation:

Note that the Real cube root is by definition the inverse of the Real cube function. That is: `root(3)(t^3) = t` and `(root(3)(t))^3 = t` for any Real value of `t`.

So given:

`f(x) = x^3+2`

`g(x) = root(3)(x-2)`

Then:

`f(g(x)) = (root(3)(x-2))^3+2 = (x-2)+2 = x`

`g(f(x)) = root(3)((x^3+2)-2) = root(3)(x^3) = x`

So `f` and `g` are mutual inverses of one another.

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Footnote

The answer above is about Real functions of Real values. The same does not hold if `f` and `g` are considered as Complex valued functions of Complex values, essentially because `t -> t^3` is not one to one as a Complex function.

For example `(-1/2+sqrt(3)/2i)^3 = 1`, so we have:

`g(f(-1/2+sqrt(3)/2i))`

`= g((-1/2+sqrt(3)/2i)^3+2)`

`= g(1+2)`

`= g(3)`

`= root(3)(3-2)`

`= root(3)(1) = 1`